Mutually permutable products of two nilpotent groups

Mutually Permutable Products of two Nilpotent Groups.

A. BALLESTER-BOLINCHES(*) - J. C. BEIDLEMAN(**) - J. COSSEY(***) H. HEINEKEN(***) - M. C. P^EDRAZA-AGUILERA(^§)

Dedicated to Professor G. Zappa on the occasion of his ninetieth birthday

1. Introduction.

Throughout the paper only finite groups are considered. Two sub- groupsAandBof a groupGare calledmutually permutableif and only if AY ˆYAandXBˆBX is true for allY BandXA; they are called totally permutable ifXY ˆYX for allYBandXA. These two con- cepts are more general than those of normal products and central pro- ducts, for instance the dihedral group of order 6 is a product of two nil- potent totally permutable subgroups. In both cases, under some restric- tions on the structure of the factors, the product is a supersoluble group (see Carocca [7] or [6, Corollary]), a class of groups considered much earlier by Zappa (see [14] and [15]). Asaad and Shaalan [1] study sufficient conditions for totally and mutually permutable products of two super- soluble subgroups to be supersoluble. More precisely they prove:

Let G be the mutually permutable product of the supersoluble sub- groups A and B. If either the product is totally permutable or A or B is nilpotent,then G is supersoluble.

(*) Indirizzo dell'A.: Departament d'AÁlgebra, Universitat de ValeÁncia, Dr.

Moliner 50, 46100 Burjassot, ValeÁncia (Spain); e-mail: Adolfo.Ballester@uv.es (**) Indirizzo dell'A.: Department of Mathematics, University of Kentucky, Lexington, Kentucky 40506-0027 U.S.A.; e-mail: clark@ms.uky.edu

(***) Indirizzo dell'A.: Department of Mathematics, Mathematical Sciences Institute, Australian National University, Canberra ACT 0200, Australia; e-mail:

John.Cossey@maths.anu.edu.au

(***) Indirizzodell'A.:MathematischesInstitut,UniversitaÈtWuÈrzburg,AmHub- land, 97074 WuÈrzburg (Germany); e-mail: heineken@mathematik.uni-wuerzburg.de (^§) Indirizzo dell'A.: Departamento de MatemaÂtica Aplicada, E.T.S.I.A, Uni- versidad PoliteÂcnica de Valencia, Camino de Vera, 46071, Valencia (Spain); e-mail:

mpedraza@mat.upv.es

The above results have been studied within the framework of formation theory. This was the beginning of an intensive study of this kind of pro- ducts, even in the infinite case (see [3,4,5,8]). Moreover Ballester-Bo- linches, Cossey and Esteban-Romero have shown that totally permutable products of nilpotent groups are extensions of an abelian by a nilpotent group (see [2, Theorem 1]). We are interested here to find more in- formation about the structure of the product of two nilpotent mutually permutable subgroups. In particular, if G is the mutually permutable product of two nilpotent subgroups of odd order, then we obtain thatGis abelian by nilpotent.

2. Results.

First of all recall that ifN denotes the class of all nilpotent groups, then the N-residual of a group G, denoted by G^N, is the intersection of all normal subgroupsNofGsuch thatG=N2 N. It is clear thatG=G^N 2 N. In the following G⁰ will denote the derived subgroup of G and G²ˆ

ˆ<g²:g2G>.

LEMMA. Let GˆAB be the mutually permutable product of the nil- potent subgroups A and B. Assume that G^N is a p-group and that G^N A.

Then G satisfies one of the following statements: (a) G²is nilpotent; (b) G^N is abelian.

PROOF. Assume the result is not true and G is a minimal counter- example, so thatGis not nilpotent,G^N 6ˆ1 andpis a prime divisor ofG;

also, by supersolubility [1],p>2. We begin with two statements and collect the consequences afterwards.

(i) O_p⁰(G)ˆ1.

The conditions of the Lemma carry over to quotient groups. Assume thatO_p⁰(G)ˆM6ˆ1:ThenM\G^N ˆ1 andMbelongs to the hypercenter ofGsinceMandMG^N=G^N are operator isomorphic. Now the nilpotence of (G=M)² leads to the nilpotence ofG²M=M and G²; on the other hand, if (G=M)^N ˆ(G^N M)=Mis abelian, so isG^N. Thus the statement is true by minimality ofG.

In particular, the Fitting subgroupF(G) ofGis ap-group andpis the biggest prime dividing the order ofG.

(ii) O^p(G)ˆG.

LetTbe any normal subgroup ofG; by [6], Lemma 1 (i),(ii) we have that (T\A)(T\B) is a normal subgroup ofGand that it is the product of the two mutually permutable subgroups T\A and T\B. Furthermore jG:(T\A)(T\B)j is a divisor of jG:Tj². Applying this to the smallest normal subgroup ofp-power indexO^p(G)ˆK;we haveKˆ(K\A)(K\B), and forK 6ˆGwe find that the Lemma is true forK. IfK²is nilpotent, so is G²ˆK²F(G); likewise K^N ˆG^N and if one is abelian, the other is, too.

Again minimality ofGyields the result.

Now we collect the consequences:F(G) is the Sylowp- subgroupPofG by (i), furthermore F(G)ˆG^N by (ii) and therefore F(G)ˆG^N ˆG⁰. Since CG(F(G))F(G) and G^N A, we obtain thatA does not possess elements of order prime top, soAˆG^N ˆPandGˆAHwhereHis the Hallp⁰- subgroup ofB. MoreoverG=AHis abelian of exponent dividing p 1.

(iii) [A;H]ˆA.

This follows fromAˆ[G;A]ˆ[H;A]A⁰[H;A]F(A).

(iv) A\BA⁰.

By [10, Proposition I(1.7), p. 206] applied to A=A⁰ and H we obtain A=A⁰ˆ([A;H]A⁰=A⁰)C_A=A⁰(H) with trivial intersection of the factors since the p-group is abelian. Now (iii) yields (B\A)A⁰=A⁰CA=A⁰(H)ˆ1 and this is (iv).

LetTˆ[A⁰;A]. We have the following strengthening of (iv):

(v) (A\B)TˆA⁰.

First, (A\B)Tis normalized byHand byA. Every subgroupU with (A\B)TUAis normalized by B sinceUˆUB\A. So Hinduces power automorphisms on A=(A\B)T. By a result of Cooper (see [9, Corollary 2.2.2, p. 339]) we have [A;[A;H]](A\B)T. Now (v) follows from (iii) and (iv).

(vi) IfAis nonabelian, thenG² is nilpotent.

Choose an elementx6ˆ1 ofH, it induces a non-identity automorphism on A=A⁰. Since all subgroups Dwith A⁰DA are normalized by Band therefore byx, there is a numberssuch thatx ¹uxA⁰ˆu^sA⁰for allu2A;

also x ¹[u₁;u₂]xTˆ[u^s₁;u^s₂]Tˆ[u₁;u₂]^s2T for all u₁;u₂2A. But x cen- tralizesA⁰=T, sos²1 mod. exp(A⁰=T) andx ¹uxA⁰ˆu ¹A⁰for allu2A.

Now [x²;A]A⁰; by a famous result of P. Hall [12, Theorem 7] we have: If the normal subgroupMof the groupTis nilpotent andT=M⁰is nilpotent,

then T is nilpotent. Applying this to A and hx²;Aiyields that hx²;Aiis nilpotent and in particular thep⁰- elementx²centralizesA. Sox²ˆ1 and H²ˆ1 sincex2Hwas arbitrary. ThereforeG²ˆAH²ˆAis nilpotent.

We have shown that there is no minimal counterexample. Now the proof is complete.

Notice that under the conditions of the Lemma the following is always true:

(+)G^N \Bˆ1 if and only ifG^N is abelian.

We are now in the position to prove our

THEOREM. Let GˆAB be the mutually permutable product of A and B with A and B nilpotent. Then(G²)^N is abelian. In particular,if G is of odd order we obtain(G^N)is abelian.

PROOF. Assume the result is not true and thatGis a minimal coun- terexample. Again the result of Asaad and Shaalan [1] yields thatGis su- persoluble. So we have thatG^N G⁰F(G) andPF(G) ifPis the Sylow p-subgroup ofGwithpthe largest prime divisor of the order ofG. Clearly, every quotient groupG=NofGinherits the hypothesis. We deduce at once

(i) Gpossesses only one minimal normal subgroup.

Assume thatN₁ andN₂ are two different minimal normal subgroups, then ((G²)^N)⁰N_iforiˆ1;2 and ((G²)^N)⁰N₁\N₂ ˆ1.

By (i), we have now that PˆF(G) and therefore G^N G⁰ PG²G.

It is a celebrated result of Wielandt (see [13, Satz 4.6, p. 676]) that in the product of two nilpotent groupsX;Ythe corresponding Hallp-subgroups permute with each other and that their product is a Hallp-subgroup ofXY, for every set of primesp. In our case we havePˆ(P\A)(P\B) and the product of the twop-complements ofAandBrespectively gives rise to a p-complementQofG, with (Q\A)(Q\B)ˆQ, andGˆPQ. The quotient groupG=PQis abelian by (i) and supersolubility ofG.

We study now the subgroup BP. Our intention is to prove that BP satisfies the hypothesis of the Lemma. We consider the intersections ofA andBwithPandQand obtain

(ii) B\QnormalizesA\P.

The subgroup A\P is the Sylow p-subgroup of (B\Q)Aˆ

ˆ(B\Q)(A\Q)(A\P)ˆQ(A\P), and (ii) follows.

(iii) BPˆ(B\Q)Pˆ(A\P)Bsatisfies the conditions of the Lem- ma, taking (A\P) andBinstead ofAandB.

Notice that [B\Q;P] is a normal subgroup of (B\Q)PˆBP. Now [B\Q;P]ˆ[B\Q;(A\P)(B\P)]ˆ[B\Q;A\P]A\Pby (ii). But also (BP)^N [B\Q;P]A\P (notice that B\Q acts on P and then Pˆ[B\Q;P]C_P(B\Q), which shows thatBP=[B\Q;P] is nilpotent). By symmetry we have also

(iv) APsatisfies the conditions of the Lemma, taking there (B\P) andAinstead ofAandB.

(v) APandBPare normal subgroups ofG.

This follows sinceG=Pis abelian.

Now we will exclude all possibilities for the counterexampleG.

(vi) (AP)^N can not be abelian.

Assume first that the normal subgroupAPofGis nilpotent. This means APF(G)ˆP andPˆA(B\P). Let Hbe the Hall p'-subgroup of B.

ThenG^N ˆ[G^N;H][P;H]ˆ[A;H]A. By our Lemma,G^N is abelian orG²is nilpotent, so also (G²)^N is abelian.

Now consider the case that (AP)^N ˆW is nontrivial and abelian. If (BP)² is nilpotent, then 1W(AP)(BP)²Gis a sequence of normal subgroups ofGwithWabelian, (AP)(BP)²=Wnilpotent, andG=(AP)(BP)² of exponent 2. So (G²)^N is abelian. If also (BP)^N is nontrivial and abelian, we have on one hand by (+) and (iv) W\Aˆ1 andW B\P, on the other hand by (+) and (iii), (BP)^N \Bˆ1 and (BP)^N A\P. Now (BP)^N \W ˆ1 contrary to the existence of only one minimal normal subgroup.

We come to the conclusion

(vii) There is no counterexample to the Theorem.

By (vi) we know that the only possibility left is that (AP)²and (BP)²are nilpotent. But then (AP)²(BP)²ˆG²is nilpotent. This finishes the proof of the Theorem.

The nilpotent residual of G need not be abelian in the situation con- sidered. In the following we give two examples:

EXAMPLE1. Consider the group Bˆ ha; bja³ˆ[b; a; a]; b³ˆ[b; a; a]²; [b; a]³ˆ[b; a; a]³ˆ[b; a; b]ˆ1i: Bhas order3⁴ and nilpotency class3.

1. All elements of BnB⁰ have order 9. Notice that (ab^k)³ˆ

ˆa³b^3k[b;a;a]^kˆ[b;a;a]^mwithmˆ1‡2k‡k1 mod o([b;a;a]). Thus B³ˆ h[b;a;a]i.

2. B⁰ˆ h[a;b];[b;a;a]i.

3. Let a be the automorphism of B given by a^aˆa², b^aˆ

ˆb²[b;a;a]²ˆb⁵. Note thatadefines an automorphism of the groupBby realizing that [a;b]^aˆ[a²;b⁵]ˆ[a ¹;b ¹] is conjugate to [a;b].

4. Let dˆ[b;a][b ¹;a ¹]. It is clear by the above remark that d^a ˆd. Therefore we can considerAˆ hd;ai ˆ hdi hai.

LetGˆ[B]haibe the corresponding semidirect product. The groupG is a mutually permutable product ofAandB. Moreover,Core_G(A\B)ˆ1 andG^N ˆBis not abelian.

EXAMPLE2. Letpbe an odd prime; denote byPan extraspecial group of orderp^2t‡1 witht1 and of exponentp. It is clear that the subgroup F(P)ˆP⁰ˆZ(P) has orderp. MoreoverP has nilpotency class2and the quotientP=Z(P) is elementary abelian. Now the application of [10, 20.8 (b), p. 81] yields that there exists an automorphismaof order2ofP such thata fixes the elements ofZ(P) and induces inversion onP=Z(P).

Consider now Gˆ[P]haithe corresponding semidirect product. De- noteBˆPandAˆ ha;Z(P)i. ThenGis the mutually permutable product of A and B both nilpotent. Moreover A\BˆZ(P) and G^N ˆP is not abelian.

Acknowledgments. The authors are indebted to the referee for hints how to improve the transparency of this paper. The first and fifth authors are supported by Proyecto MTM2004-08219-C02-02 (MCyT) and FEDER (European Union).

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Manoscritto pervenuto in redazione il 9 gennaio 2006.